course Mth 272 017.
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Given Solution: `a First we factor x the demoninator: 3/ [ x(x-3) ] Use partial fractions: 3/ [ x(x-3) ] = A/x + B/(x-3) Multiply both sides by common denominator x(x-3) to get 3 = A(x-3) + B(x) or 3 = (A+B) x - 3 A, which is the same as 0 x + 3 = (A + B) x - 3 A. The coefficients of x on both sides must be the same so we have A + B = 0 (coefficients of x) and -3 A = 3 . From the second we get A = -1. Substituting this into the first we solve to get B = 1. So our integrand is 3 / (x^2 - 3x) = -1 / (x-3) + 1 / x. The integration is straightforward. We get ln |x-3| - ln |x| + c , which we rewrite using the laws of logarithms as ln | (x-3) / x | + c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: correct ********************************************* Question: `qQuery problem 6.3.29 (was 6.3.27) integrate (x+2) / (x^2 - 4x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A/x + B/ (x-4) zeros= 0 and 4 x+2= A(x-4) + B(x) plug in 0 for x..........-4A=2 A=-½ plug in 4 for x.........4B=6 B=3/2 ANSWER (-1/2)/x + (3/2)/(x-4) confidence rating: 3. This was another joy to work. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a We factor x out of the denominator to get (x+2)/ [ x(x-4) ] Use partial fractions: (x+2)/x(x-4) = A/x + B/(x-4) Multiply both sides by common denominator: x+2 = A(x-4) + B(x) or x+2 = (A+B) x - 4 A. Thus A + B = 1 and -4 A = 2 so A = -1/2 and B = 3/2. Our integrand becomes (-1/2) / x + (3/2) / (x-4). The general antiderivative is easily found to be 3/2 ln |x-4| - 1/2 ln |x| + c, which can be expressed as 1/2 ln ( |x-4|^3 / | x | ) + c &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): correct ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " xxxx